Factor completely. $16x^2-81=$
Solution: Both $16x^2$ and $81$ are perfect squares, since $16x^2=({4x})^2$ and $81=({9})^2$. $16x^2-81 = ({4x})^2-({9})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={4x}$ and ${b}={9}$ : $({4x})^2 - ({9})^2 =({4x}+{9})({4x}-{9})$ In conclusion, $16x^2-81=(4x+9)(4x-9)$ Remember that you can always check your factorization by expanding it.